Abstract
Most models of DNA transposition invoke replication of the transposable element, but it is not clear whether a 'co-integrate' is an obligatory intermediate in the pathway leading to the production of simple insertions during transposition. Such an intermediate can be accounted for only by a replicative transposition scheme. Bacteriophage Mu is a temperate phage that can either lysogenize or lyse its host, and it encodes at least two modes of transposition as judged by the end-products generated by the process. During the lytic development of the integrated prophage, co-integrates are the predominant end-products; transposition is coupled to replication during this phase. A small number of simple insertions are also produced during the lytic growth, but during transposition from the infecting phage into the host chromosome, simple insertions are the main end-products. Conditions can be found where the choice between the two kinds of end-products depends on a delicate balance between the essential transposition functions encoded by Mu. Experiments have suggested that the simple insertions which arise during transposition from the infecting phage may do so without Mu DNA replication. Here I demonstrate using an infecting phage with completely methylated DNA, a dam- (DNA adenine methylase) host and a combination of restriction enzymes that can cut either fully methylated or unmethylated DNA but not hemi-methylated DNA, that transposition of the phage DNA into the host chromosome does not involve a duplication of its DNA. This result may also have significance for other transposons that do not appear to go through a co-integrate intermediate during transposition.
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